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CHXII03:ELECTROCHEMISTRY

330066 The oxidation potentials of $Z n, C u, A g, H_{2} a n d N i$ are $0 .76, - 0 .34, - 0 .80, 0 .00, 0 .25$ volt, respectively. Which of the following reactions will provide maximum voltage?

1

H_{2} + N i^{2 +} \to 2 H^{+} + Ni

2

H_{2} + C u^{2 +} \to 2 H^{+} + Cu

3

Zn + 2A g^{+} \to 2Ag + Z n^{2 +}

4

Zn + C u^{2 +} \to Cu + Z n^{2 +}

Explanation:

$E_{c e l l}^{o} = 0 .8 - (- 0 .76) = 1 .56 V$
Maximum voltage is produced when electrode having minimum oxidation potential is made cathode and maximum oxidation potential is made anode.

CHXII03:ELECTROCHEMISTRY

330067 If the half-cell reactions are given as
$I . F e^{2 +} (a q) + 2 e^{-} \to F e (s), E^{\circ} = - 0.44 V$
$I I . 2 H^{+} (a q) + \frac{1}{2} O_{2} (g) + 2 e^{-} \to H_{2} O (l),$
$E^{\circ} = + 1.23 V$
then $E^{\circ}$ for the reaction,
$F e (s) + 2 H^{+} + \frac{1}{2} O_{2} (g) \to F e^{2 +} (a q) + H_{2} O (l)$

1

+ 1.67 V

2

- 1.67 V

3

+ 0.79 V

4

- 0.79 V

Explanation:

$E {^{\circ}}_{c e l l} = E {^{\circ}}_{c a t h o d e} - E {^{\circ}}_{a n o d e}$
$= + 1.23 - (- 0.44)$
$= + 1.67 V$

CHXII03:ELECTROCHEMISTRY

330068 The EMF of the cell $Ni | N i^{2 +} | C u^{2 +} | Cu (s)$ is 0.59 volt. The standard reduction electrode potential of copper electrode is 0.34 volt. The standard reduction electrode potential of nickel electrode will be

1 –0.025 volt

2 –0.50 volt

3 0.25 volt

4 –0.25 volt

Explanation:

$E_{C u^{2 +} / C u}^{o} {- E}_{N i^{+ 2} / N i}^{o} {= E}_{c e l l}^{o}$
$0 .34 - x = 0 .59$
$x = 0 .25 V$

CHXII03:ELECTROCHEMISTRY

330069 The $E^{o}$ for the following cell is 0.34 V. $I n (s) | I n (O H)_{3} (a q) | {S b O}_{2}^{-} (a q) | S b (s)$
[Using $E^{o} = - 1 .0 V$ for the $I n (O H_{3}) |$ In couple, calculate $E^{o}$ for the ${S b O}_{2}^{-} | S b$ half-reaction:]

1

- 1 .34

2

+ 0 .66

3

+ 0.82

4

- 0.66

Explanation:

$0 {.34 = 1 + E}_{{S b O}_{2}^{-} | S b}^{o}$
$E_{{S b O}_{2}^{-} | S b}^{o} = 0 .34 - 1 = - 0 .66 V$

CHXII03:ELECTROCHEMISTRY

330070 The standard electrode potential for the half cell reactions are
$Z n^{2 +} + 2 e^{-} \to Z n; E^{o} = - 0.76 V$
$F e^{2 +} + 2 e^{-} \to F e; E^{o} = - 0.44 V$
The e.m.f of the cell reaction
$F e^{2 +} + Z n \to Z n^{2 +} + F e$

1

- 1 .20 V

2

+ 1 .20 V

3

+ 0 .32 V

4

- 0 .32 V

Explanation:

Given cell reaction is
$Z n + F e^{2 +} \to Z n^{2 +} + F e$
In the above reaction, Zn is being oxidised and $F e^{2 +}$ is being reduced, thus the cell can berepresented as,
$Z n | Z n^{2 +} | F e^{2 +} | F e$
$E_{c e l l} = E_{R i g h t} - E_{L e f t} = E_{F e^{2} | F e} - E_{Z n^{2 +} | Z n}$
$= (- 0 .44) - (- 0 .76)$
$= 0 .76 - 0 .44 = 0 .32 V$

KCET - 2009

CHXII03:ELECTROCHEMISTRY

330066 The oxidation potentials of $Z n, C u, A g, H_{2} a n d N i$ are $0 .76, - 0 .34, - 0 .80, 0 .00, 0 .25$ volt, respectively. Which of the following reactions will provide maximum voltage?

1

H_{2} + N i^{2 +} \to 2 H^{+} + Ni

2

H_{2} + C u^{2 +} \to 2 H^{+} + Cu

3

Zn + 2A g^{+} \to 2Ag + Z n^{2 +}

4

Zn + C u^{2 +} \to Cu + Z n^{2 +}

Explanation:

$E_{c e l l}^{o} = 0 .8 - (- 0 .76) = 1 .56 V$
Maximum voltage is produced when electrode having minimum oxidation potential is made cathode and maximum oxidation potential is made anode.

CHXII03:ELECTROCHEMISTRY

330067 If the half-cell reactions are given as
$I . F e^{2 +} (a q) + 2 e^{-} \to F e (s), E^{\circ} = - 0.44 V$
$I I . 2 H^{+} (a q) + \frac{1}{2} O_{2} (g) + 2 e^{-} \to H_{2} O (l),$
$E^{\circ} = + 1.23 V$
then $E^{\circ}$ for the reaction,
$F e (s) + 2 H^{+} + \frac{1}{2} O_{2} (g) \to F e^{2 +} (a q) + H_{2} O (l)$

1

+ 1.67 V

2

- 1.67 V

3

+ 0.79 V

4

- 0.79 V

Explanation:

$E {^{\circ}}_{c e l l} = E {^{\circ}}_{c a t h o d e} - E {^{\circ}}_{a n o d e}$
$= + 1.23 - (- 0.44)$
$= + 1.67 V$

CHXII03:ELECTROCHEMISTRY

330068 The EMF of the cell $Ni | N i^{2 +} | C u^{2 +} | Cu (s)$ is 0.59 volt. The standard reduction electrode potential of copper electrode is 0.34 volt. The standard reduction electrode potential of nickel electrode will be

1 –0.025 volt

2 –0.50 volt

3 0.25 volt

4 –0.25 volt

Explanation:

$E_{C u^{2 +} / C u}^{o} {- E}_{N i^{+ 2} / N i}^{o} {= E}_{c e l l}^{o}$
$0 .34 - x = 0 .59$
$x = 0 .25 V$

CHXII03:ELECTROCHEMISTRY

330069 The $E^{o}$ for the following cell is 0.34 V. $I n (s) | I n (O H)_{3} (a q) | {S b O}_{2}^{-} (a q) | S b (s)$
[Using $E^{o} = - 1 .0 V$ for the $I n (O H_{3}) |$ In couple, calculate $E^{o}$ for the ${S b O}_{2}^{-} | S b$ half-reaction:]

1

- 1 .34

2

+ 0 .66

3

+ 0.82

4

- 0.66

Explanation:

$0 {.34 = 1 + E}_{{S b O}_{2}^{-} | S b}^{o}$
$E_{{S b O}_{2}^{-} | S b}^{o} = 0 .34 - 1 = - 0 .66 V$

CHXII03:ELECTROCHEMISTRY

330070 The standard electrode potential for the half cell reactions are
$Z n^{2 +} + 2 e^{-} \to Z n; E^{o} = - 0.76 V$
$F e^{2 +} + 2 e^{-} \to F e; E^{o} = - 0.44 V$
The e.m.f of the cell reaction
$F e^{2 +} + Z n \to Z n^{2 +} + F e$

1

- 1 .20 V

2

+ 1 .20 V

3

+ 0 .32 V

4

- 0 .32 V

Explanation:

Given cell reaction is
$Z n + F e^{2 +} \to Z n^{2 +} + F e$
In the above reaction, Zn is being oxidised and $F e^{2 +}$ is being reduced, thus the cell can berepresented as,
$Z n | Z n^{2 +} | F e^{2 +} | F e$
$E_{c e l l} = E_{R i g h t} - E_{L e f t} = E_{F e^{2} | F e} - E_{Z n^{2 +} | Z n}$
$= (- 0 .44) - (- 0 .76)$
$= 0 .76 - 0 .44 = 0 .32 V$

KCET - 2009

CHXII03:ELECTROCHEMISTRY

330066 The oxidation potentials of $Z n, C u, A g, H_{2} a n d N i$ are $0 .76, - 0 .34, - 0 .80, 0 .00, 0 .25$ volt, respectively. Which of the following reactions will provide maximum voltage?

1

H_{2} + N i^{2 +} \to 2 H^{+} + Ni

2

H_{2} + C u^{2 +} \to 2 H^{+} + Cu

3

Zn + 2A g^{+} \to 2Ag + Z n^{2 +}

4

Zn + C u^{2 +} \to Cu + Z n^{2 +}

Explanation:

$E_{c e l l}^{o} = 0 .8 - (- 0 .76) = 1 .56 V$
Maximum voltage is produced when electrode having minimum oxidation potential is made cathode and maximum oxidation potential is made anode.

CHXII03:ELECTROCHEMISTRY

330067 If the half-cell reactions are given as
$I . F e^{2 +} (a q) + 2 e^{-} \to F e (s), E^{\circ} = - 0.44 V$
$I I . 2 H^{+} (a q) + \frac{1}{2} O_{2} (g) + 2 e^{-} \to H_{2} O (l),$
$E^{\circ} = + 1.23 V$
then $E^{\circ}$ for the reaction,
$F e (s) + 2 H^{+} + \frac{1}{2} O_{2} (g) \to F e^{2 +} (a q) + H_{2} O (l)$

1

+ 1.67 V

2

- 1.67 V

3

+ 0.79 V

4

- 0.79 V

Explanation:

$E {^{\circ}}_{c e l l} = E {^{\circ}}_{c a t h o d e} - E {^{\circ}}_{a n o d e}$
$= + 1.23 - (- 0.44)$
$= + 1.67 V$

CHXII03:ELECTROCHEMISTRY

330068 The EMF of the cell $Ni | N i^{2 +} | C u^{2 +} | Cu (s)$ is 0.59 volt. The standard reduction electrode potential of copper electrode is 0.34 volt. The standard reduction electrode potential of nickel electrode will be

1 –0.025 volt

2 –0.50 volt

3 0.25 volt

4 –0.25 volt

Explanation:

$E_{C u^{2 +} / C u}^{o} {- E}_{N i^{+ 2} / N i}^{o} {= E}_{c e l l}^{o}$
$0 .34 - x = 0 .59$
$x = 0 .25 V$

CHXII03:ELECTROCHEMISTRY

330069 The $E^{o}$ for the following cell is 0.34 V. $I n (s) | I n (O H)_{3} (a q) | {S b O}_{2}^{-} (a q) | S b (s)$
[Using $E^{o} = - 1 .0 V$ for the $I n (O H_{3}) |$ In couple, calculate $E^{o}$ for the ${S b O}_{2}^{-} | S b$ half-reaction:]

1

- 1 .34

2

+ 0 .66

3

+ 0.82

4

- 0.66

Explanation:

$0 {.34 = 1 + E}_{{S b O}_{2}^{-} | S b}^{o}$
$E_{{S b O}_{2}^{-} | S b}^{o} = 0 .34 - 1 = - 0 .66 V$

CHXII03:ELECTROCHEMISTRY

330070 The standard electrode potential for the half cell reactions are
$Z n^{2 +} + 2 e^{-} \to Z n; E^{o} = - 0.76 V$
$F e^{2 +} + 2 e^{-} \to F e; E^{o} = - 0.44 V$
The e.m.f of the cell reaction
$F e^{2 +} + Z n \to Z n^{2 +} + F e$

1

- 1 .20 V

2

+ 1 .20 V

3

+ 0 .32 V

4

- 0 .32 V

Explanation:

Given cell reaction is
$Z n + F e^{2 +} \to Z n^{2 +} + F e$
In the above reaction, Zn is being oxidised and $F e^{2 +}$ is being reduced, thus the cell can berepresented as,
$Z n | Z n^{2 +} | F e^{2 +} | F e$
$E_{c e l l} = E_{R i g h t} - E_{L e f t} = E_{F e^{2} | F e} - E_{Z n^{2 +} | Z n}$
$= (- 0 .44) - (- 0 .76)$
$= 0 .76 - 0 .44 = 0 .32 V$

KCET - 2009

CHXII03:ELECTROCHEMISTRY

330066 The oxidation potentials of $Z n, C u, A g, H_{2} a n d N i$ are $0 .76, - 0 .34, - 0 .80, 0 .00, 0 .25$ volt, respectively. Which of the following reactions will provide maximum voltage?

1

H_{2} + N i^{2 +} \to 2 H^{+} + Ni

2

H_{2} + C u^{2 +} \to 2 H^{+} + Cu

3

Zn + 2A g^{+} \to 2Ag + Z n^{2 +}

4

Zn + C u^{2 +} \to Cu + Z n^{2 +}

Explanation:

$E_{c e l l}^{o} = 0 .8 - (- 0 .76) = 1 .56 V$
Maximum voltage is produced when electrode having minimum oxidation potential is made cathode and maximum oxidation potential is made anode.

CHXII03:ELECTROCHEMISTRY

330067 If the half-cell reactions are given as
$I . F e^{2 +} (a q) + 2 e^{-} \to F e (s), E^{\circ} = - 0.44 V$
$I I . 2 H^{+} (a q) + \frac{1}{2} O_{2} (g) + 2 e^{-} \to H_{2} O (l),$
$E^{\circ} = + 1.23 V$
then $E^{\circ}$ for the reaction,
$F e (s) + 2 H^{+} + \frac{1}{2} O_{2} (g) \to F e^{2 +} (a q) + H_{2} O (l)$

1

+ 1.67 V

2

- 1.67 V

3

+ 0.79 V

4

- 0.79 V

Explanation:

$E {^{\circ}}_{c e l l} = E {^{\circ}}_{c a t h o d e} - E {^{\circ}}_{a n o d e}$
$= + 1.23 - (- 0.44)$
$= + 1.67 V$

CHXII03:ELECTROCHEMISTRY

330068 The EMF of the cell $Ni | N i^{2 +} | C u^{2 +} | Cu (s)$ is 0.59 volt. The standard reduction electrode potential of copper electrode is 0.34 volt. The standard reduction electrode potential of nickel electrode will be

1 –0.025 volt

2 –0.50 volt

3 0.25 volt

4 –0.25 volt

Explanation:

$E_{C u^{2 +} / C u}^{o} {- E}_{N i^{+ 2} / N i}^{o} {= E}_{c e l l}^{o}$
$0 .34 - x = 0 .59$
$x = 0 .25 V$

CHXII03:ELECTROCHEMISTRY

330069 The $E^{o}$ for the following cell is 0.34 V. $I n (s) | I n (O H)_{3} (a q) | {S b O}_{2}^{-} (a q) | S b (s)$
[Using $E^{o} = - 1 .0 V$ for the $I n (O H_{3}) |$ In couple, calculate $E^{o}$ for the ${S b O}_{2}^{-} | S b$ half-reaction:]

1

- 1 .34

2

+ 0 .66

3

+ 0.82

4

- 0.66

Explanation:

$0 {.34 = 1 + E}_{{S b O}_{2}^{-} | S b}^{o}$
$E_{{S b O}_{2}^{-} | S b}^{o} = 0 .34 - 1 = - 0 .66 V$

CHXII03:ELECTROCHEMISTRY

330070 The standard electrode potential for the half cell reactions are
$Z n^{2 +} + 2 e^{-} \to Z n; E^{o} = - 0.76 V$
$F e^{2 +} + 2 e^{-} \to F e; E^{o} = - 0.44 V$
The e.m.f of the cell reaction
$F e^{2 +} + Z n \to Z n^{2 +} + F e$

1

- 1 .20 V

2

+ 1 .20 V

3

+ 0 .32 V

4

- 0 .32 V

Explanation:

Given cell reaction is
$Z n + F e^{2 +} \to Z n^{2 +} + F e$
In the above reaction, Zn is being oxidised and $F e^{2 +}$ is being reduced, thus the cell can berepresented as,
$Z n | Z n^{2 +} | F e^{2 +} | F e$
$E_{c e l l} = E_{R i g h t} - E_{L e f t} = E_{F e^{2} | F e} - E_{Z n^{2 +} | Z n}$
$= (- 0 .44) - (- 0 .76)$
$= 0 .76 - 0 .44 = 0 .32 V$

KCET - 2009

CHXII03:ELECTROCHEMISTRY

330066 The oxidation potentials of $Z n, C u, A g, H_{2} a n d N i$ are $0 .76, - 0 .34, - 0 .80, 0 .00, 0 .25$ volt, respectively. Which of the following reactions will provide maximum voltage?

1

H_{2} + N i^{2 +} \to 2 H^{+} + Ni

2

H_{2} + C u^{2 +} \to 2 H^{+} + Cu

3

Zn + 2A g^{+} \to 2Ag + Z n^{2 +}

4

Zn + C u^{2 +} \to Cu + Z n^{2 +}

Explanation:

$E_{c e l l}^{o} = 0 .8 - (- 0 .76) = 1 .56 V$
Maximum voltage is produced when electrode having minimum oxidation potential is made cathode and maximum oxidation potential is made anode.

CHXII03:ELECTROCHEMISTRY

330067 If the half-cell reactions are given as
$I . F e^{2 +} (a q) + 2 e^{-} \to F e (s), E^{\circ} = - 0.44 V$
$I I . 2 H^{+} (a q) + \frac{1}{2} O_{2} (g) + 2 e^{-} \to H_{2} O (l),$
$E^{\circ} = + 1.23 V$
then $E^{\circ}$ for the reaction,
$F e (s) + 2 H^{+} + \frac{1}{2} O_{2} (g) \to F e^{2 +} (a q) + H_{2} O (l)$

1

+ 1.67 V

2

- 1.67 V

3

+ 0.79 V

4

- 0.79 V

Explanation:

$E {^{\circ}}_{c e l l} = E {^{\circ}}_{c a t h o d e} - E {^{\circ}}_{a n o d e}$
$= + 1.23 - (- 0.44)$
$= + 1.67 V$

CHXII03:ELECTROCHEMISTRY

330068 The EMF of the cell $Ni | N i^{2 +} | C u^{2 +} | Cu (s)$ is 0.59 volt. The standard reduction electrode potential of copper electrode is 0.34 volt. The standard reduction electrode potential of nickel electrode will be

1 –0.025 volt

2 –0.50 volt

3 0.25 volt

4 –0.25 volt

Explanation:

$E_{C u^{2 +} / C u}^{o} {- E}_{N i^{+ 2} / N i}^{o} {= E}_{c e l l}^{o}$
$0 .34 - x = 0 .59$
$x = 0 .25 V$

CHXII03:ELECTROCHEMISTRY

330069 The $E^{o}$ for the following cell is 0.34 V. $I n (s) | I n (O H)_{3} (a q) | {S b O}_{2}^{-} (a q) | S b (s)$
[Using $E^{o} = - 1 .0 V$ for the $I n (O H_{3}) |$ In couple, calculate $E^{o}$ for the ${S b O}_{2}^{-} | S b$ half-reaction:]

1

- 1 .34

2

+ 0 .66

3

+ 0.82

4

- 0.66

Explanation:

$0 {.34 = 1 + E}_{{S b O}_{2}^{-} | S b}^{o}$
$E_{{S b O}_{2}^{-} | S b}^{o} = 0 .34 - 1 = - 0 .66 V$

CHXII03:ELECTROCHEMISTRY

330070 The standard electrode potential for the half cell reactions are
$Z n^{2 +} + 2 e^{-} \to Z n; E^{o} = - 0.76 V$
$F e^{2 +} + 2 e^{-} \to F e; E^{o} = - 0.44 V$
The e.m.f of the cell reaction
$F e^{2 +} + Z n \to Z n^{2 +} + F e$

1

- 1 .20 V

2

+ 1 .20 V

3

+ 0 .32 V

4

- 0 .32 V

Explanation:

Given cell reaction is
$Z n + F e^{2 +} \to Z n^{2 +} + F e$
In the above reaction, Zn is being oxidised and $F e^{2 +}$ is being reduced, thus the cell can berepresented as,
$Z n | Z n^{2 +} | F e^{2 +} | F e$
$E_{c e l l} = E_{R i g h t} - E_{L e f t} = E_{F e^{2} | F e} - E_{Z n^{2 +} | Z n}$
$= (- 0 .44) - (- 0 .76)$
$= 0 .76 - 0 .44 = 0 .32 V$

KCET - 2009